Question

For the reversible reaction, $A+2B\to 3C$, if equilibrium constant is $1\times {10}^2$ at 27°C then the value of $\Delta r{G}^{\circ }$ at the same temperature will be:
$(R=2cal{K}^{-1}mo{l}^{-1})$

• A. $-8.21kcalmo{l}^{-1}$
• B. $-2.76kcalmo{l}^{-1}$
• C. $-4.31kcalmo{l}^{-1}$
• D. $12.27kcalmo{l}^{-1}$

Answer 1

The correct option is B $-2.76kcalmo{l}^{-1}$

Equilibrium constant, $\left(K\right)=1\times {10}^2$

$\Delta r{G}^{\circ }=-RTlnK$

$=-2.303RTlogK$

$=-2.303\times 2\times 300\times log{10}^2$

$=-2.303\times 2\times 300\times 2$

$=-2.76kcalmo{l}^{-1}$